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Topic: The Twin Paradox !...What's That All About ? (Read 12574 times)

Indeed, but ONLY the 'travelling' twin would be able to measure an acceleration using, say, an inertial accelerometer. The twin on the earth would not 'experience' a measureable acceleration, and this is why you can't equate the two frames of reference. This is where the asymmetry creeps into the paradox and why the travelling twin really does return younger than the stay-at-home. During any steady-velocity part of the journey both twins would observe the clocks of the other twin to be running slow - this is a symmetrical effect and applies regardless of the direction of travel

There is another affect that we need to remember i.e. In a strong gravitational field, general relativity says that clocks run slower than in a weak gravitational field. So if you really want to make this impressive, send teh travelling twin to orbit close to a black hole for a couple of months!

Uncle Albert equates gravity and acceleration, doesn't he? They are precisely the same thing; Newton2.

My name is Albert too.

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A little acceleration for a long time or a lot for a little time will both have the effect of changing the RATE of the clocks on the traveller's ship for ever, once the engines are turned off.

I don't have the possibility to post a simple drawing of space-time diagrams, however you can find many on the web. If I find one I'll post it. You would notice that the amount of acceleration/deceleration and the interval of time it lasts, is not so important, in the evaluation of the age difference of the twins, if they move apart for a long distance and at high relative speed. Infact, what really counts is exactly that: distance travelled and relative speeds.

Imagine an orthogonal cartesian ref. frame; you put time on the vertical axis and distance x on the horizontal.

In the ref. frame of A: T is the time needed by B to reach Proxima Centaury (5 years, as you remember).L is earth-Proxima Centaury distance (4 light years).A vertical segment from (0,0) to (0,2T) represent the world-line of A;a couple of bent segments, the first from (0,0) to (L,T), the second from (L,T) to (0,2T) represent the world-line of B. The proper-time of a twin is Δs/c where Δs is the space-time interval between two events in a world line.So, A's proper time is Δs/c evaluated in A's world-line, from (0,0) to (0,2T). Since Δs = Sqrt[(cΔt)2 - (Δx)2] we have: Δs(A) = Sqrt[(c*2T)2 - 0] = 2cT. Δs(B) = 2Sqrt[(c*T)2 - (L)2]Let's evaluate the difference of proper times:Δs(A)/c - Δs(B)/c = 2T - 2Sqrt[(T)2 - (L/c)2] == 2*5 - 2Sqrt[(5)2 - (4)2] == 10 - 2Sqrt(9) = 10 - 6 = 4 yars.As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.

Minkowski recast Einstein's version of Special Relativity (SR) on a new stage, Minkowski spacetime. The Twin Paradox has a very simple resolution in this framework. The crucial concept is the proper time of a moving body.

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lyner

As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.The fact is that he DID accelerate, and did not stay in an inertial frame, - which is what got him to the speed with which you are doing your calculations. It is only the fact that he accelerated that makes him different from the 'stationary' twin. You cannot argue with that, I am sure. Without an accelerometer, neither of the twins would, initially, know whether it was he or his brother or both of them who was 'moving away' (trains moving out of stations etc.). That was the start of the original idea of SR.As far as I can see, your calculations are, of course, fine but the link in my last post evaluates how the acceleration has produced precisely the sort effect that your calculations show. This is not surprising; there are often many ways of showing a result / killing a cat. However - the link takes more into account and doesn't rely on any talk of a fixed reference frame or a journey to any particular star. It just describes my simple two twin situation - out in the middle of nowhere - and determines which of them ages slower.

As you see, B accelerates around point (L,T) but how much it accelerates doesn't show in the evaluation.The fact is that he DID accelerate, and did not stay in an inertial frame, - which is what got him to the speed with which you are doing your calculations. It is only the fact that he accelerated that makes him different from the 'stationary' twin. You cannot argue with that, I am sure.

Yes, even because I've already said it several times in my posts!

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Without an accelerometer, neither of the twins would, initially, know whether it was he or his brother or both of them who was 'moving away' (trains moving out of stations etc.). That was the start of the original idea of SR.

Of course.

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As far as I can see, your calculations are, of course, fine but the link in my last post evaluates how the acceleration has produced precisely the sort effect that your calculations show. This is not surprising; there are often many ways of showing a result / killing a cat. However - the link takes more into account and doesn't rely on any talk of a fixed reference frame or a journey to any particular star. It just describes my simple two twin situation - out in the middle of nowhere - and determines which of them ages slower.

If you consider the space-time diagram of a twins paradox type experiment, you can see that the twins form a triangle.

Acceleration means that the triangle isn't perfect, that the points of the triangle are actually slightly rounded by the acceleration.

However, this can be made to be a very small effect, by taking the limit of high accelerations.

What the twin paradox says is very similar to the geometrical theorem that the shortest distance between two points is a straight line, or eqivalently, the "triangle inequality" that says that the sum of the length of two sides of a triangle is always longer than the remaining third side.

In the case of the twins paradox, though, the statement becomes that the observer following a geodesic path (analogous to the straight line, especially in the flat geometry of SR) has the longest proper time. The time is the longest, but the distance in the Euclidean geometrical analogy is the shortest. This has to do with the difference between Euclidean geometry with it's ++++ metric signature, and the Lorentzian geomery of SR with it's -+++ signature.

Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

That is why people say that the twin paradox isn't about the rounding of the corners due to the acceleration - it's due to the angle between the observers. This angle on a space-time diagram is simply a change of velocity. What is important is not the rate at which the velocity changes (the accleration) , but what is often called "delta-V", the change in velocity.

The geometrical analogy of the "twins pardox" would be the "triangle paradox". The triangle paradox would say:

If I go directly from A to B, I travel the shortest distance. And if I go from A to C to B, I travel a longer distance. It's "paradoxical" (?) that when I go from A to C to B, that this distance is longer than going straight from A to B. Furthermore, if I look at the distance from A to B, that's shorter than going from A to C to B. Now AB is the shortest distance!

Of course people gemerally don't get confused by the "triangle paradox" - it's really not that much more difficult not to get confused by the "twin paradox" either.

So, I repeat that:1. The Twin Paradox Does arises from the fact that one of them have to accelerate.2. The age difference has Nothing to do with the amount of acceleration.3. There is no need of GR to explain it and to make all the necessary computations.

By the way, there isn't even the need of accelerometers to understand who was the one who has accelerated!

If they send each other light pulses at costant rate (in their own ref frame), for example, one light pulse every second, after the complete journey we can understand who was the one who has accelerated just from the distribution of light pulses every twin has received from the other one.

One of them (1) sees a low frequency of pulses for longer time and a high frequency for a shorter time.The other (2) sees a low frequency of pulses for a shorter time and a high frequency for a longer time (or the two intervals of time are the same, in the limit of infinite deceleration/acceleration at the return point).

The one who has accelerated is (2).

Twin Paradox has always fascinated me!Alberto.

« Last Edit: 15/04/2007 19:38:19 by lightarrow »

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lyner

I can see that we are agreeing about all of this apart from the fact that you say the acceleration is irrelevant. If it weren't for the fact that one was accelerating, there would be no difference in velocity so no effect.

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Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

The velocity difference gives you the ANGLES in your triangle. which is what gives you something to calculate with. If you had instantaneous speed change then you would have sharp corners, which would look pretty but is not practical. Velocity is merely the integral over time of acceleration - in practice, giving curves, not sharp corners. This detail (curved or sharp) is not really relevant.Your 'delta v' is a factor from outside SR and accounts for the angle.The only paradox is there when one ignores this.The total change in velocity, of course, depends on the acceleration and how long it's accelerating for (the same as impulse (force times time) can be used in conventional dynamics problems. I can't see why this presents you with a difficulty. During the time our twin is changing velocity, it is not in an inertial frame. The reason behind all your calculations is what happens whilst it is in a non- inertial frame. The only situation in which you can ignore GR is when they are coasting past each other, without any velocity change. If you have acceleration , you can't ignore GR. But why should you want to?

I can see that we are agreeing about all of this apart from the fact that you say the acceleration is irrelevant.

Sophie, "irrelevant" for what? Did you understand what I mean when I say this?I've tried to explain this several times. This is another time. I said that:The Amount of Acceleration can be made Irrelevant for computing The amount of The age difference.Not, that Acceleration is irrelevant!

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If it weren't for the fact that one was accelerating, there would be no difference in velocity so no effect.

Of Course!

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Acceleration (the curvature at the tips of the triangle) really has very little to do with this geometrical result.

The velocity difference gives you the ANGLES in your triangle. which is what gives you something to calculate with. If you had instantaneous speed change then you would have sharp corners, which would look pretty but is not practical. Velocity is merely the integral over time of acceleration - in practice, giving curves, not sharp corners. This detail (curved or sharp) is not really relevant.Your 'delta v' is a factor from outside SR and accounts for the angle.The only paradox is there when one ignores this.The total change in velocity, of course, depends on the acceleration and how long it's accelerating for (the same as impulse (force times time) can be used in conventional dynamics problems. I can't see why this presents you with a difficulty. During the time our twin is changing velocity, it is not in an inertial frame. The reason behind all your calculations is what happens whilst it is in a non- inertial frame.

See up.

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The only situation in which you can ignore GR is when they are coasting past each other, without any velocity change. If you have acceleration , you can't ignore GR. But why should you want to?

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?ds2 = (cdt)2 - (dx)2is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.I repeat, the paradox DOES arise from the fact one of them is accelerating.But GR IS NOT NECESSARY (= we don't need to use or know that theory) to make all the considerations and calculations to solve the paradox..

lyner

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?ds2 = (cdt)2 - (dx)2is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.

Your geometry is impeccable and it gives the right result from the point of view of the twin who stays behind because HE is in an inertial frame. There is nothing wrong with your construction to show what happens on the straight bits, but you are including changes in velocity at the corners, which, as I keep saying, involves the traveler not being in an inertial frame all the time. This is where the GR bit comes in. It is this effect on him that makes him observe the one who stays behind as being older than him when he returns. It is the traveler who experiences the time-rate changes at the corners. If you don't allow for this then what is the difference between the observations of the two? There is no reference grid, so why, apart from the acceleration factor, will the experiences not be symmetrical and paradoxical? Go through the same procedure for the travelling twin and obey the rules; he is not in flat space time all the 'time' so he must see things differently; you can't use your Minowski metric in the same way for him. But this is just going round in circles and that, btw, is not an inertial frame, either!

Sorry, Sophie, but *where exactly* did I use GR to make my calculations?ds2 = (cdt)2 - (dx)2is the Minkowsky metric, valid only in a "Flat" space-time, that is, in SR and NOT in GR.

Your geometry is impeccable and it gives the right result from the point of view of the twin who stays behind because HE is in an inertial frame. There is nothing wrong with your construction to show what happens on the straight bits, but you are including changes in velocity at the corners, which, as I keep saying, involves the traveler not being in an inertial frame all the time. This is where the GR bit comes in. It is this effect on him that makes him observe the one who stays behind as being older than him when he returns. It is the traveler who experiences the time-rate changes at the corners.

What I have colored in blue is not true, and I showed you why.

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If you don't allow for this then what is the difference between the observations of the two? There is no reference grid, so why, apart from the acceleration factor, will the experiences not be symmetrical and paradoxical? Go through the same procedure for the travelling twin and obey the rules; he is not in flat space time all the 'time' so he must see things differently; you can't use your Minowski metric in the same way for him. But this is just going round in circles and that, btw, is not an inertial frame, either!

Let's put it in this way:The Twin Paradox CAN be explained using General Relativity.Have I ever claimed the contrary?No. I have said that this "Is not necessary".

Let's say that we have n objects, each of them moving with different velocities and accelerations in a region of space far from any massive object.Can I find their relative aging using General Relativity?YES.Can I find it using Special Relativity?YES.How is that possible?I choose an inertial reference frame and I draw space-time diagrams in that frame; then I make all the necessary computations. Does this inertial ref. frame coincide with one of the object's ref frame, as in the twin prdx? Yes or not, it doesn't matter.Why should I prefer this reference frame and not another one?Because computations are simpler.

Let's say you are in the accelerating rocket, and you want to make all the necessary computations to understand if it's you or your twin who will be younger when you meet again.The fact it's you who is accelerating, you can understand:1. with accelerometers, or2. studying the distribution of light signals arriving to you from your twin.

Then how can you make the computations? You can choose:General RelativityORSpecial Relativity.

If you draw diagrams using your ref. frame, you have to use GR; if you choose an inertial ref frame (any!) to draw your diagrams, you can use SR. You are not forced to use GR only because you are not in an inertial ref. frame, do you know what I mean? Because you don't have the necessity to use your ref. frame in your notebook's diagrams!

« Last Edit: 18/04/2007 16:04:13 by lightarrow »

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lyner

You are sounding quite convincing now. I'm all in favour of using the simplest method for getting an answer and I have some chance of following SR in my brain. BUT (and there's got to be a but) it seems that, using your argument, no one need ever use GR to explain anything. You imply that one can always explain the effects of general relativity by referring your measurements to some inertial frame, elsewhere. This can't be true for all cases, or they wouldn't have bothered to invent GR. Can we, possibly, reconcile our differences with the following scenario?Your traveling twin goes near a huge dead star, which he cannot see, because it is a black dwarf. The gravitational field of this object introduces an acceleration (we, surely, must use GR in this case, mustn't we?), which he measures and which he takes into account when he analyses his observations of his non-traveling brother. He attributes his relativistic effects to acceleration due to his engines and does your calculations , making the appropriate corrections etc. to decide how much younger he will be at the end of the experiment.Can he possibly still get the correct answer about his relative age after this experience?If he can, then I can't see why GR should ever be necessary. If he can't, then what is the difference between the acceleration due to his engines and the acceleration due to the gravitational field that he enters?I am not being argumentative for the sake of it now; I am genuinely mystified . Is the traveller actually capable of making all the corrections without taking GR into account? Unless he makes allowances for his time dilation (due to acceleration) can he, in fact, determine where he is and how fast he's going? I guess you will say "yes".Have you an example of a similar type of experiment where GR would HAVE to be used, so I can get the picture properly?

I notice that not many other people are getting involved with this.Is there anyone else who can resolve this?

You are sounding quite convincing now. I'm all in favour of using the simplest method for getting an answer and I have some chance of following SR in my brain. BUT (and there's got to be a but) it seems that, using your argument, no one need ever use GR to explain anything. You imply that one can always explain the effects of general relativity by referring your measurements to some inertial frame, elsewhere. This can't be true for all cases, or they wouldn't have bothered to invent GR. Can we, possibly, reconcile our differences with the following scenario?Your traveling twin goes near a huge dead star, which he cannot see, because it is a black dwarf. The gravitational field of this object introduces an acceleration (we, surely, must use GR in this case, mustn't we?), which he measures and which he takes into account when he analyses his observations of his non-traveling brother. He attributes his relativistic effects to acceleration due to his engines and does your calculations , making the appropriate corrections etc. to decide how much younger he will be at the end of the experiment.Can he possibly still get the correct answer about his relative age after this experience?

Good consideration. My compliments to you! I really don't know the answer. I imagine that, if the dead star is far, non rotating, and the travelling twin doesn't approach it much and so he experiences only uniform accelerations inside the rocket, then it's possible to find at least one inertial ref. frame and to draw space-time diagrams, and make all the computations necessary, so that GR is not necessary even in this case, but I'm not sure.

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If he can, then I can't see why GR should ever be necessary. If he can't, then what is the difference between the acceleration due to his engines and the acceleration due to the gravitational field that he enters?

In general, in a curved region of space, is not (always?) possible to find an inertial ref. frame, so all that could be impossible of course. I don't know much about it however.

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I am not being argumentative for the sake of it now; I am genuinely mystified.

Me too!

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Is the traveller actually capable of making all the corrections without taking GR into account? Unless he makes allowances for his time dilation (due to acceleration) can he, in fact, determine where he is and how fast he's going? I guess you will say "yes".

As I said, maybe it should be possible, in some very simple cases; certainly not in all of them.

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Have you an example of a similar type of experiment where GR would HAVE to be used, so I can get the picture properly?

In a curved region of space, not asymptotically minkowskian, with rotating massive objects, I'm pretty sure GR is necessary!

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I notice that not many other people are getting involved with this.Is there anyone else who can resolve this?

I hope so!And I hope we two will make other interesting discussions like this in other threads, in future.Bye!

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